Consider the free non-abelian group on the twenty-six letters of the alphabet with all relations of the...

language/english/pronunciation/homophone/trivial

Problem

Consider the free non-abelian group on the twenty-six letters of the
alphabet with all relations of the form <word1> = <word2>, where <word1>
and <word2> are homophones (i.e. they sound alike but are spelled
differently).  Show that every letter is trivial.

For example, be = bee, so e is trivial.

Show solution

Using Webster's Third International as a reference:

be = bee ==> e is trivial;
ail = ale ==> i is trivial;
week = weak ==> a is trivial;
lie = lye ==> y is trivial;
to = too ==> o is trivial;
two = to ==> w is trivial;
hour = our ==> h is trivial;
faggot = fagot ==> g is trivial;
bowl = boll ==> l is trivial;
gel = jel ==> j is trivial;
you = ewe ==> u is trivial;
damn = dam ==> n is trivial;
limb = limn ==> b is trivial;
bass = base ==> s is trivial;
cede = seed ==> c is trivial;
knead = need ==> k is trivial;
add = ad ==> d is trivial;
awful = offal ==> f is trivial;
mama = mamma ==> m is trivial;
grip = grippe ==> p is trivial;
cue = queue ==> q is trivial;
carrel = carol ==> r is trivial;
butt = but ==> t is trivial;
lox = locks ==> x is trivial;
faze = phase ==> z is trivial;
vlei = fly ==> v is trivial.

For a related problem, see _The Jimmy's Book_ (_The American Mathematical
Monthly_, Vol. 93, Num. 8 (Oct. 1986), p.  637):

Consider the free group on twenty-six letters A, ..., Z.  Mod out by
the relation that defines two words to be equivalent if (a) one is a
permutation of the other and (b) each appears as a legitimate English
word in the dictionary.  Identify the center of this group.

    -- clong@cnj.digex.com (Chris Long)